How do you solve $ln (x + 6) + ln (x - 6) = 0$ $ln(x+6)+ln(x-6)=0$ ?

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How do you solve $ln (x + 6) + ln (x - 6) = 0$ $ln(x+6)+ln(x-6)=0$ ?

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Answer 1 · 2015-07-08T23:38:52

I found: $x = \sqrt{37} = 6.082$ $x=sqrt(37)=6.082$

Explanation:

You can start by using the rule of logs:
$log a + log b = log (a \cdot b)$ $loga+logb=log(a*b)$

In your case you get:
$ln [(x + 6) \cdot (x - 6)] = 0$ $ln[(x+6)*(x-6)]=0$

Now you can use the definition of log as:
${log}_{a} x = b \to x = a^{b}$ $log_ax=b -> x=a^b$ remembering that the natural log is: $ln x = {log}_{e} x$ $lnx=log_ex$

so:
$(x + 6) (x - 6) = e^{0}$ $(x+6)(x-6)=e^0$
$(x + 6) (x - 6) = 1$ $(x+6)(x-6)=1$
rearranging:
$x^{2} - 36 = 1$ $x^2-36=1$
$x^{2} = 37$ $x^2=37$
$x = \pm \sqrt{37} = \pm 6.082$ $x=+-sqrt(37)=+-6.082$
The negative $x$ $x$ cannot be accepted (substituted back it gives a negative argument in the original logs).

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How do you solve $ln (x + 6) + ln (x - 6) = 0$ $ln(x+6)+ln(x-6)=0$ ?

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How do you solve ln(x+6)+ln(x−6)=0ln(x+6)+ln(x-6)=0?

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How do you solve $ln (x + 6) + ln (x - 6) = 0$ $ln(x+6)+ln(x-6)=0$ ?